OFFSET
1,1
COMMENTS
This sequence is a generalization of the 4th problem proposed for the pupils in grade 6 during the 19th Mathematical Festival at Moscow in 2008.
Some notations: s = side of the tiled square, a = side of small squares, b = side of large squares, and z = number of small squares = number of large squares.
Side s of such tiled squares must satisfy the Diophantine equation s^2 = z * (a^2+b^2).
If q is a term, k * q is another term for k > 1.
REFERENCES
Ivan Yashchenko, Invitation to a Mathematical Festival, pp. 10 and 102, MSRI, Mathematical Circles Library, 2013.
EXAMPLE
-> Example of type 1:
Square 10 x 10 with a = 1, b = 2, s = 10, z = 20.
___ ___ _ ___ ___ _
| | |_| | |_|
|___|___|_|___|___|_|
| | |_| | |_| with 10 elementary 2 x 5 rectangles
|___|___|_|___|___|_|
| | |_| | |_| ___ ___ _
|___|___|_|___|___|_| | | |_|
| | |_| | |_| |___|___|_|
|___|___|_|___|___|_|
| | |_| | |_|
|___|___|_|___|___|_|
.
-> Example of type 2:
Square 15 x 15 with a = 3, b = 4, s = 15, z = 9.
________ ________ ________ _____
| | | | |
| | | | |
| | | |_____|
|_______ |________|________| |
| | | | |
| | | |_____|
| | | | |
|________|________|________| |
| | | |_____|
| | | | |
| | | | |
|_____ __|___ ____|_ ______|_____|
| | | | | |
| | | | | |
|_____|______|______|______|_____|
Remarks:
- With terms as 10, 20, ... we only obtain sides of squares of type 1:
10 is a term of this type because the square 10 X 10 only can be tiled with 20 squares of size 1 X 1 and 20 squares of size 2 X 2 (see first example),
20 is another term of this type because the square 20 X 20 only can be tiled with 80 squares of size 1 x 1 and 80 squares of size 2 x 2.
- With terms as 15, 65, ... we only obtain sides of squares of type 2:
15 is a term of this type because the square 15 X 15 only can be tiled with 9 squares of size 3 X 3 and 9 squares of size 4 X 4 (see second example),
65 is another term of this type because the square 65 X 65 only can be tiled with 25 squares of size 5 X 5 and 25 squares of size 12 X 12.
- With terms as 30, 60, ... we obtain both sides of squares of type 1 and of type 2:
30 is a term of type 1 because the square 30 X 30 can be tiled with 180 squares of size 1 X 1 and 180 squares of size 2 X 2, but,
30 is also a term of type 2 because the square 30 X 30 can be tiled with 9 squares of size 6 X 6 and 9 squares of size 8 X 8.
PROG
(PARI) pts(lim) = my(v=List(), m2, s2, h2, h); for(middle=4, lim-1, m2=middle^2; for(small=1, middle, s2=small^2; if(issquare(h2=m2+s2, &h), if(h>lim, break); listput(v, [small, middle, h])))); vecsort(Vec(v)); \\ A009000
isdp4(s) = my(k=1, x); while(((x=k^4 - (k-1)^4) <= s), if (x == s, return (1)); k++); return(0);
isokp2(s) = {if (!isdp4(s), return(0)); if (s % 2, my(vp = pts(s)); for (i=1, #vp, my(vpi = vp[i], a = vpi[1], b = vpi[2], c = vpi[3]); if (a*c/(c-b) == s, return(1)); ); ); }
isok2(s) = {if (isokp2(s), return (1)); fordiv(s, d, if ((d>1) || (d<s), if (isokp2(s/d), return (1)))); }
isokp1(s) = {if (!(s % 2) && ispower(s/2, 4), return (0)); my(m = sqrtnint(s, 3)); vecsearch(setbinop((x, y)->x*y*(x^2+y^2), [1..m]), s); }
isok1(s) = {if (isokp1(s), return (1)); fordiv(s, d, if ((d>1) || (d<s), if (isokp1(s/d), return (1)))); }
isok(s) = isok1(s) || isok2(s); \\ Michel Marcus, Jun 04 2021
CROSSREFS
KEYWORD
nonn
AUTHOR
Bernard Schott, May 15 2021
EXTENSIONS
Corrected by Michel Marcus, May 18 2021
Incorrect term 145 removed by Michel Marcus, Jun 04 2021
STATUS
approved